Electrostatic Potential and Capacitance

Step: Find the amount of energy required to bring the charges closer.
Electrostatic potential energy between two charges separated by distance \(r\) is given by;
\(U = \frac{kq_1q_2}{r}\) 
Initial electrostatic potential energy between the charges is given by;
\(U_i = \frac{9\times 10^9 \times 1.1 \times 10^{-8}\times 1.1\times 10 ^{-9}}{0.1} = 10.89 \times 10^{-7}~\text{J}\)
The final electrostatic potential energy between the charges is given by;
\(U_f = \frac{9\times 10^9 \times 1.1 \times 10^{-8}\times 1.1\times 10 ^{-9}}{0.01} = 10.89 \times 10^{-6}~\text{J}\)
The amount of energy required to bring the charges closer is given by;
\(\Rightarrow W = U_f-U_i = 10.89\times10^{-6}[1-0.1] \)
\(\Rightarrow W = 10.89\times10^{-6}\times0.9 \approx 10^{-5}~\text{J}\)
Hence, option (4) is the correct answer